What is the solution for y^{''}+y=csc^2(x) ?

The solution for y^{''}+y=csc^2(x) is y=c_{1}cos(x)+c_{2}sin(x)-cos(x)ln(tan(x/2))-1

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y^{''}+y=csc^2(x)

y^{^{''}} + y = csc^{2} (x)

y = c_{1} cos (x) + c_{2} sin (x) - cos (x) ln (tan (\frac{x}{2})) - 1

Solution steps

y^{''} (x) + y = csc^{2} (x)

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) - cos (x) ln (tan (\frac{x}{2})) - 1

y = c_{1} cos (x) + c_{2} sin (x) - cos (x) ln (tan (\frac{x}{2})) - 1

\int x (x^{2} + 1)^{4} d x

e^{x} - 4 x^{3}

\frac{d}{d x} (\frac{x - 1}{x + 1})

\frac{\partial}{\partial x} (\frac{1}{2} ln (x + y))

y^{^{''}} + k^{2} y = 0

What is the solution for y^{''}+y=csc^2(x) ?
The solution for y^{''}+y=csc^2(x) is y=c_{1}cos(x)+c_{2}sin(x)-cos(x)ln(tan(x/2))-1